Extraneous Solutions In Rational Equations Solving (2m)/(2m+3) - (2m)/(2m-3) = 1

In mathematics, extraneous solutions are solutions that arise during the process of solving an equation but do not satisfy the original equation. These solutions often occur when dealing with rational equations, radical equations, or equations involving absolute values. Identifying and eliminating extraneous solutions is a crucial step in obtaining accurate results. In this comprehensive guide, we will delve into the process of solving the equation $\frac{2m}{2m+3} - \frac{2m}{2m-3} = 1$ and meticulously determine the number of extraneous solutions. We will provide a step-by-step approach, ensuring a clear understanding of each stage, and highlight the importance of verifying solutions to avoid errors.

Understanding Extraneous Solutions

Before diving into the solution, it's essential to grasp the concept of extraneous solutions. Extraneous solutions are values that emerge as solutions during the algebraic manipulation of an equation but do not satisfy the original equation. This often happens when we perform operations that are not reversible, such as squaring both sides of an equation or multiplying both sides by an expression that could be zero. For rational equations, extraneous solutions typically arise when a solution makes the denominator of any fraction in the original equation equal to zero, leading to an undefined expression. Therefore, it is imperative to check all potential solutions in the original equation to identify and discard any extraneous solutions.

Why Extraneous Solutions Occur

• Non-Reversible Operations: Operations like squaring both sides of an equation can introduce extraneous solutions. For example, if we have $\sqrt{x} = -2$, squaring both sides gives $x = 4$, but substituting $x = 4$ back into the original equation yields $\sqrt{4} = 2$, not -2. Thus, $x = 4$ is an extraneous solution.
• Rational Equations: In rational equations, extraneous solutions can occur when a potential solution makes the denominator zero. Division by zero is undefined, so any value that results in a zero denominator must be excluded. For instance, in the equation $\frac{1}{x-1} = 0$, if we were to find $x = 1$ as a solution, it would be an extraneous solution because it makes the denominator zero.

The Importance of Verification

The process of solving an equation is incomplete without verifying the solutions. Verification involves substituting each potential solution back into the original equation to ensure it holds true. This step is critical in identifying and eliminating extraneous solutions. By verifying, we guarantee that the solutions we obtain are valid and accurately reflect the original equation's conditions. This practice not only ensures accuracy but also deepens our understanding of the problem and the solution process.

Solving the Equation (2m)/(2m+3) - (2m)/(2m-3) = 1

To solve the given equation, $\frac{2m}{2m+3} - \frac{2m}{2m-3} = 1$, we will follow a systematic approach. This involves finding a common denominator, simplifying the equation, and then solving for $m$. Each step is crucial in arriving at the correct solutions. However, it's important to be vigilant for potential extraneous solutions along the way.

Step 1: Finding a Common Denominator

The first step in solving the equation is to find a common denominator for the fractions. The denominators in the equation are $2m + 3$ and $2m - 3$. The least common denominator (LCD) is the product of these two expressions, which is $(2m + 3)(2m - 3)$. We need to rewrite each fraction with this common denominator:

• Multiply the first fraction, $\frac{2m}{2m+3}$, by $\frac{2m-3}{2m-3}$:

  $$\frac{2m}{2m+3} \cdot \frac{2m-3}{2m-3} = \frac{2m(2m-3)}{(2m+3)(2m-3)}$$

• Multiply the second fraction, $\frac{2m}{2m-3}$, by $\frac{2m+3}{2m+3}$:

  $$\frac{2m}{2m-3} \cdot \frac{2m+3}{2m+3} = \frac{2m(2m+3)}{(2m+3)(2m-3)}$$

Now, rewrite the original equation with the common denominator:

$$\frac{2m(2m-3)}{(2m+3)(2m-3)} - \frac{2m(2m+3)}{(2m+3)(2m-3)} = 1$$

Step 2: Combining the Fractions

With a common denominator, we can combine the fractions by subtracting the numerators:

$$\frac{2m(2m-3) - 2m(2m+3)}{(2m+3)(2m-3)} = 1$$

Expand the terms in the numerator:

$$\frac{4m^2 - 6m - (4m^2 + 6m)}{(2m+3)(2m-3)} = 1$$

Simplify the numerator by combining like terms:

$$\frac{4m^2 - 6m - 4m^2 - 6m}{(2m+3)(2m-3)} = 1$$

$$\frac{-12m}{(2m+3)(2m-3)} = 1$$

Step 3: Simplifying the Denominator

Next, simplify the denominator by multiplying the binomials $(2m + 3)(2m - 3)$. This is a difference of squares, so the result is $(2m)^2 - (3)^2$:

$$\frac{-12m}{4m^2 - 9} = 1$$

Step 4: Solving for m

To solve for $m$, we'll first multiply both sides of the equation by the denominator $4m^2 - 9$:

$$-12m = 4m^2 - 9$$

Rearrange the equation to form a quadratic equation by moving all terms to one side:

$$4m^2 + 12m - 9 = 0$$

Now, we can solve this quadratic equation. We can use the quadratic formula, which is given by:

$$m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

In our equation, $a = 4$, $b = 12$, and $c = -9$. Plugging these values into the quadratic formula, we get:

$$m = \frac{-12 \pm \sqrt{12^2 - 4(4)(-9)}}{2(4)}$$

$$m = \frac{-12 \pm \sqrt{144 + 144}}{8}$$

$$m = \frac{-12 \pm \sqrt{288}}{8}$$

Simplify the square root by factoring out the largest perfect square:

$$m = \frac{-12 \pm \sqrt{144 \cdot 2}}{8}$$

$$m = \frac{-12 \pm 12\sqrt{2}}{8}$$

Divide all terms by 4:

$$m = \frac{-3 \pm 3\sqrt{2}}{2}$$

So, we have two potential solutions:

$$m_1 = \frac{-3 + 3\sqrt{2}}{2}$$

$$m_2 = \frac{-3 - 3\sqrt{2}}{2}$$

Identifying Extraneous Solutions

Now that we have our potential solutions, the crucial step is to check for extraneous solutions. To do this, we substitute each value of $m$ back into the original equation and verify if it holds true. We also need to ensure that these values do not make the denominators in the original equation equal to zero.

Step 1: Checking for Zero Denominators

Before substituting the values into the original equation, let's identify the values of $m$ that would make the denominators zero. The denominators in the original equation are $2m + 3$ and $2m - 3$. Setting each equal to zero:

• $2m + 3 = 0$ gives $m = -\frac{3}{2}$
• $2m - 3 = 0$ gives $m = \frac{3}{2}$

Our potential solutions, $m_1 = \frac{-3 + 3\sqrt{2}}{2}$ and $m_2 = \frac{-3 - 3\sqrt{2}}{2}$, are not equal to $\pm \frac{3}{2}$, so they do not make the denominators zero. This is a good initial check, but we still need to verify them in the original equation.

Step 2: Substituting Potential Solutions into the Original Equation

We will substitute each potential solution, $m_1$ and $m_2$, into the original equation: $\frac{2m}{2m+3} - \frac{2m}{2m-3} = 1$.

For $m_1 = \frac{-3 + 3\sqrt{2}}{2}$:

Substitute $m_1$ into the equation:

$$\frac{2(\frac{-3 + 3\sqrt{2}}{2})}{2(\frac{-3 + 3\sqrt{2}}{2})+3} - \frac{2(\frac{-3 + 3\sqrt{2}}{2})}{2(\frac{-3 + 3\sqrt{2}}{2})-3} = 1$$

Simplify the expression:

$$\frac{-3 + 3\sqrt{2}}{-3 + 3\sqrt{2} + 3} - \frac{-3 + 3\sqrt{2}}{-3 + 3\sqrt{2} - 3} = 1$$

$$\frac{-3 + 3\sqrt{2}}{3\sqrt{2}} - \frac{-3 + 3\sqrt{2}}{-6 + 3\sqrt{2}} = 1$$

$$\frac{-1 + \sqrt{2}}{\sqrt{2}} - \frac{-3 + 3\sqrt{2}}{-6 + 3\sqrt{2}} = 1$$

$$\frac{-\sqrt{2} + 2}{2} - \frac{-1 + \sqrt{2}}{-2 + \sqrt{2}} = 1$$

$$\frac{-\sqrt{2} + 2}{2} - \frac{(-1 + \sqrt{2})(-2 - \sqrt{2})}{(-2 + \sqrt{2})(-2 - \sqrt{2})} = 1$$

$$\frac{-\sqrt{2} + 2}{2} - \frac{2 + \sqrt{2} - 2\sqrt{2} - 2}{4 - 2} = 1$$

$$\frac{-\sqrt{2} + 2}{2} - \frac{-\sqrt{2}}{2} = 1$$

$$\frac{-\sqrt{2} + 2 + \sqrt{2}}{2} = 1$$

$$\frac{2}{2} = 1$$

$$1 = 1$$

So, $m_1 = \frac{-3 + 3\sqrt{2}}{2}$ is a valid solution.

For $m_2 = \frac{-3 - 3\sqrt{2}}{2}$:

Substitute $m_2$ into the equation:

$$\frac{2(\frac{-3 - 3\sqrt{2}}{2})}{2(\frac{-3 - 3\sqrt{2}}{2})+3} - \frac{2(\frac{-3 - 3\sqrt{2}}{2})}{2(\frac{-3 - 3\sqrt{2}}{2})-3} = 1$$

Simplify the expression:

$$\frac{-3 - 3\sqrt{2}}{-3 - 3\sqrt{2} + 3} - \frac{-3 - 3\sqrt{2}}{-3 - 3\sqrt{2} - 3} = 1$$

$$\frac{-3 - 3\sqrt{2}}{-3\sqrt{2}} - \frac{-3 - 3\sqrt{2}}{-6 - 3\sqrt{2}} = 1$$

$$\frac{-1 - \sqrt{2}}{-\sqrt{2}} - \frac{-3 - 3\sqrt{2}}{-6 - 3\sqrt{2}} = 1$$

$$\frac{\sqrt{2} + 2}{2} - \frac{-1 - \sqrt{2}}{-2 - \sqrt{2}} = 1$$

$$\frac{\sqrt{2} + 2}{2} - \frac{(-1 - \sqrt{2})(-2 + \sqrt{2})}{(-2 - \sqrt{2})(-2 + \sqrt{2})} = 1$$

$$\frac{\sqrt{2} + 2}{2} - \frac{2 - \sqrt{2} + 2\sqrt{2} - 2}{4 - 2} = 1$$

$$\frac{\sqrt{2} + 2}{2} - \frac{\sqrt{2}}{2} = 1$$

$$\frac{\sqrt{2} + 2 - \sqrt{2}}{2} = 1$$

$$\frac{2}{2} = 1$$

$$1 = 1$$

So, $m_2 = \frac{-3 - 3\sqrt{2}}{2}$ is also a valid solution.

Step 3: Determining the Number of Extraneous Solutions

Both potential solutions, $m_1 = \frac{-3 + 3\sqrt{2}}{2}$ and $m_2 = \frac{-3 - 3\sqrt{2}}{2}$, satisfy the original equation. Therefore, there are no extraneous solutions in this case.

Conclusion

In summary, we solved the equation $\frac{2m}{2m+3} - \frac{2m}{2m-3} = 1$ by finding a common denominator, simplifying the equation, and using the quadratic formula to find potential solutions. We then meticulously checked these solutions in the original equation and found that both $m_1 = \frac{-3 + 3\sqrt{2}}{2}$ and $m_2 = \frac{-3 - 3\sqrt{2}}{2}$ are valid. Therefore, there are no extraneous solutions. This comprehensive step-by-step approach ensures accuracy and highlights the importance of verifying solutions to avoid errors. Understanding the concept of extraneous solutions and how to identify them is crucial in solving rational equations effectively.

Final Answer

The final answer is A. 0.